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Posts published in “Geometry”

花花酱 LeetCode 391. Perfect Rectangle

Given N axis-aligned rectangles where N > 0, determine if they all together form an exact cover of a rectangular region.

Each rectangle is represented as a bottom-left point and a top-right point. For example, a unit square is represented as [1,1,2,2]. (coordinate of bottom-left point is (1, 1) and top-right point is (2, 2)).

Example 1:

Example 2:

Example 3:

Example 4:

Solution: Counting corner points

C++

花花酱 LeetCode 986. Interval List Intersections

Given two lists of closed intervals, each list of intervals is pairwise disjoint and in sorted order.

Return the intersection of these two interval lists.

(Formally, a closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b.  The intersection of two closed intervals is a set of real numbers that is either empty, or can be represented as a closed interval.  For example, the intersection of [1, 3] and [2, 4] is [2, 3].)

Example 1:

Input: A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
Reminder: The inputs and the desired output are lists of Interval objects, and not arrays or lists.

Note:

  1. 0 <= A.length < 1000
  2. 0 <= B.length < 1000
  3. 0 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9

Solution: Two pointers

Time complexity: O(m + n)
Space complexity: O(1)

C++

Python3

花花酱 LeetCode 973. K Closest Points to Origin

We have a list of points on the plane.  Find the K closest points to the origin (0, 0).

(Here, the distance between two points on a plane is the Euclidean distance.)

You may return the answer in any order.  The answer is guaranteed to be unique (except for the order that it is in.)

Example 1:

Input: points = [[1,3],[-2,2]], K = 1 
Output: [[-2,2]]
Explanation: The distance between (1, 3) and the origin is sqrt(10). The distance between (-2, 2) and the origin is sqrt(8). Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin. We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].

Example 2:

Input: points = [[3,3],[5,-1],[-2,4]], K = 2 
Output: [[3,3],[-2,4]] (The answer [[-2,4],[3,3]] would also be accepted.)

Note:

  1. 1 <= K <= points.length <= 10000
  2. -10000 < points[i][0] < 10000
  3. -10000 < points[i][1] < 10000

Solution: Sort

Time complexity: O(nlogn)
Space complexity: O(n)

C++

Python3

花花酱 LeetCode 963. Minimum Area Rectangle II

Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes.

If there isn’t any rectangle, return 0.

Example 1:

Input: [[1,2],[2,1],[1,0],[0,1]]
Output: 2.00000
Explanation: The minimum area rectangle occurs at [1,2],[2,1],[1,0],[0,1], with an area of 2.

Example 2:

Input: [[0,1],[2,1],[1,1],[1,0],[2,0]]
Output: 1.00000
Explanation: The minimum area rectangle occurs at [1,0],[1,1],[2,1],[2,0], with an area of 1.

Example 3:

Input: [[0,3],[1,2],[3,1],[1,3],[2,1]]
Output: 0
Explanation: There is no possible rectangle to form from these points.

Example 4:

Input: [[3,1],[1,1],[0,1],[2,1],[3,3],[3,2],[0,2],[2,3]]
Output: 2.00000
Explanation: The minimum area rectangle occurs at [2,1],[2,3],[3,3],[3,1], with an area of 2.

Note:

  1. 1 <= points.length <= 50
  2. 0 <= points[i][0] <= 40000
  3. 0 <= points[i][1] <= 40000
  4. All points are distinct.
  5. Answers within 10^-5 of the actual value will be accepted as correct.

Solution: HashTable

Iterate all possible triangles and check the opposite points that creating a quadrilateral.

Time complexity: O(n^3)
Space complexity: O(n)

C++

花花酱 LeetCode 939. Minimum Area Rectangle

Problem

Given a set of points in the xy-plane, determine the minimum area of a rectangle formed from these points, with sides parallel to the x and y axes.

If there isn’t any rectangle, return 0.

Example 1:

Input: [[1,1],[1,3],[3,1],[3,3],[2,2]]
Output: 4

Example 2:

Input: [[1,1],[1,3],[3,1],[3,3],[4,1],[4,3]]
Output: 2

Note:

  1. 1 <= points.length <= 500
  2. 0 <= points[i][0] <= 40000
  3. 0 <= points[i][1] <= 40000
  4. All points are distinct.

Solution 1: HashTable + Brute Force

Try all pairs of points to form a diagonal and see whether pointers of another diagonal exist or not.

Assume two points are (x0, y0), (x1, y1) x0 != x1 and y0 != y1. The other two points will be (x0, y1), (x1, y0)

Time complexity: O(n^2)

Space complexity: O(n)

C++

C++ / 1D hashtable